What is the Change of Length for a Hanging Steel Cable?

[brainslush]

Homework Statement

A rope (steel cable) is hanging in an empty hole. What is the change of length?
L = 9km
E_{steel} = 2 · 1011 N/m2
p_{steel} = 7.8 · 103 kg/m3

Homework Equations

The Attempt at a Solution

Well, I thought that the cable could also be considered as a spring. I found in another thread that the change of length of a hanging spring is
<br /> k \,dx = \bigg( \frac{l}{L_0} \bigg) g \, dm<br />

<br /> dx = \frac{gl}{L_0 k} \frac{M}{L_0} \, dl<br />

<br /> dx = \frac{Mgl}{{L_0}^2 k} \, dl<br />

<br /> x = \frac{Mg}{{L_0}^2 k} \int_0^{L_0} l \, dl<br />

<br /> x = \frac{Mg}{{L_0}^2 k} \frac{{L_0}^2}{2}<br />

<br /> x = \frac{Mg}{2k}<br />

I understood this part so far.

but I'm not sure whether my reconstruction is correct.

x = \frac{g\rho_{steel}L^{2}}{2E_{steel}}

 

[Beaker87]

To solve this problem, use the formula for Young's modulus, which can be rearranged to read:
\Delta L = \frac {F L_0}{A_0 E}
where A₀ is the cross-sectional area.To solve this for a cable with a continuous mass distribution, the formula becomes:
\Delta L = \int_0^{L_0} \frac{FdL}{A_0E}The force is equal to the density times the volume, or:
F = \rho A_0 L_0 gWhen integrating this upward from the bottom of the cable, the force term becomes:
F = \rho A_0 L g
where L is the amount of cable below the point in question.Substituting this into the formula for a cable of continuous mass distribution:
\Delta L = \int_0^{L_0} \frac{\rho L g}{E}dL
the A₀ terms, of course, cancel each other out.And finally, integrating gives us:
\Delta L = \frac{\rho {L_0}^2 g}{2E}

The same as your reconstruction.

I hope this helps,
Beaker87